1. Field of the Invention
The present invention relates to a current mirror circuit, and more particularly, to a current mirror circuit which can operate at a low voltage, and can output a current equal to the input current even if the constituent transistors have a relatively small current-amplification factor .beta..
2. Description of the Related Art
FIG. 1 attached hereto shows a typical current mirror circuit hitherto known, which is comprised of NPN transistors T1 and T2. If transistors T1 and T2 had the same current-amplification factor .beta. which is infinitely great, the collector current of transistor T2, i.e., the output current Iout of the current mirror circuit, should be equal to the input current Iin supplied to the collector of transistor T1. In effect, since the current-amplification factor .beta. of NPN transistors T1, T2 is limited, the output current Iout depends on .beta. and is less than the input current Iin, as is evident from the following equations: EQU Iout=Ic2=Ic1=Iin/[1+(2/.beta..sub.N)]
where Ic1 is the collector current of transistor I1, Ib1 is the base current of transistor T1, Ic2 is the collector current of transistor T2, and .beta..sub.N is the emitter-ground current-amplification factor of transistors T1 and T2. When .beta..sub.N is 350, 70, 20, and 10, Iout and Iin will have the following relationship: EQU Iout=0.9943 Iin (.beta..sub.N =350) EQU Iout=0.9722 Iin (.beta..sub.N =70) EQU Iout=0.9091 Iin (.beta..sub.N =20) EQU Iout=0.8333 Iin (.beta..sub.N =10)
Obviously, the ratio of Iout to Iin greatly depends upon the current-amplification factor .beta. of transistor T1 and T2.
FIG. 2 shows an improved current mirror circuit, wherein the ratio of Iout to Iin is less dependent on the current amplification factor .beta. of transistor T1 and T2. This circuit differs from the circuit of FIG. 1 in that NPN transistor T3 is coupled between the collector of transistor T1 and the base of transistor T2. Since the collector current of transistor T3 is supplied to the bases of transistors T1 and T2, the ratio of Iout to Iin is less dependent upon .beta.. More specifically, in the circuit shown in FIG. 2, Iout is given as: EQU Iout=Iin[1+(2/.beta..sub.N.sup.2)]
Therefore: EQU Iout=0.9999 Iin (.beta..sub.N =350) EQU Iout=0.9996 Iin (.beta..sub.N =70) EQU Iout=0.9950 Iin (.beta..sub.N =20) EQU Iout=0.9804 Iin (.beta..sub.N =10)
Although the ratio of Iout to Iin is considerably less dependent on .beta. than in the circuit of FIG. 1, the circuit of FIG. 2 can not operate so efficiently at a low voltage as the current mirror circuit shown in FIG. 1. This is because the input-terminal voltage of the circuit shown in FIG. 2 is 1.4 V, i.e., the sum of the base-emitter voltage Vbe1 of transistor T1 and the base-emitter voltage Vbe3 of transistor T3, whereas the input-terminal voltage of the circuit shown in FIG. 1 is only 0.7 V, i.e., the base-emitter voltage Vbe1 of transistor T1.